NCERT Class 12 Physics — Magnetism and Matter, Chapter 5 Exercises. All 7 questions solved.

The seven exercises at the end of Chapter 5 are all about the magnetic dipole. Questions 5.1, 5.2 and 5.5 put a bar magnet in a uniform field and ask for the torque on it, its potential energy, and the work needed to turn it. Questions 5.3, 5.4 and 5.6 do the same for a current-carrying solenoid, which first has to be recognised as a dipole and given a magnetic moment. Question 5.7 turns round and asks for the field the dipole itself produces. Three results cover all of it:

• Magnetic moment of a solenoid — $m = NIA$ for $N$ turns of area $A$ carrying current $I$, directed along the axis by the right-hand rule.
• Dipole in a uniform field — torque $\vec\tau = \vec m\times\vec B$, of magnitude $mB\sin\theta$, and potential energy $U = -\vec m\cdot\vec B = -mB\cos\theta$, where $\theta$ is the angle between $\vec m$ and $\vec B$.
• Field of a short bar magnet at a distance $r$ from its centre — $B = \dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}$ on the axis and $B = \dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}$ on the equatorial line, with $\dfrac{\mu_0}{4\pi} = 10^{-7}\ \text{T m A}^{-1}$.

Key insight. For every question here, a bar magnet and a current-carrying solenoid are the same object: a magnetic dipole, completely described by its moment $\vec m$. Once $m$ is known — given outright (5.1, 5.2, 5.5, 5.7) or worked out as $NIA$ (5.3, 5.4, 5.6) — the torque, the energy and the field all follow from $m$, $B$ and the angle between them.

Question 5.1

A short bar magnet placed with its axis at $30^\circ$ with a uniform external magnetic field of $0.25\ \text{T}$ experiences a torque of magnitude equal to $4.5\times10^{-2}\ \text{J}$. What is the magnitude of magnetic moment of the magnet?

Solution. The magnet’s moment points along its axis, so the angle between $\vec m$ and $\vec B$ is $30^\circ$, and the torque is $\tau = mB\sin\theta$. Turning this round for $m$:

$$m = \frac{\tau}{B\sin\theta} = \frac{4.5\times10^{-2}}{(0.25)\sin30^\circ} = \frac{4.5\times10^{-2}}{0.125} = 0.36\ \text{J T}^{-1}$$

The book quotes the torque in joules. A joule and a newton metre are the same combination of base units, though torque is normally written in $\text{N m}$.

Question 5.2

A short bar magnet of magnetic moment $m = 0.32\ \text{J T}^{-1}$ is placed in a uniform magnetic field of $0.15\ \text{T}$. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Question 5.2 (a)

Solution. The torque $mB\sin\theta$ vanishes only at $\theta = 0^\circ$ and $\theta = 180^\circ$, so those are the two equilibrium positions. Equilibrium is stable where the potential energy $U = -mB\cos\theta$ is a minimum, which is at $\theta = 0^\circ$, with $\vec m$ along $\vec B$. Turn the magnet a little away from there and the torque turns it back.

$$U = -mB\cos0^\circ = -(0.32)(0.15) = -4.8\times10^{-2}\ \text{J}$$

Question 5.2 (b)

Solution. At $\theta = 180^\circ$, with $\vec m$ pointing against $\vec B$, the potential energy is a maximum. The torque is still zero there, but the slightest disturbance produces a torque that turns the magnet further away, towards the stable position, so the equilibrium is unstable.

$$U = -mB\cos180^\circ = +(0.32)(0.15) = +4.8\times10^{-2}\ \text{J}$$

Question 5.3

A closely wound solenoid of $800$ turns and area of cross section $2.5\times10^{-4}\ \text{m}^2$ carries a current of $3.0\ \text{A}$. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Solution. Why it is like a bar magnet. Each turn of the solenoid is a small current loop, and a current loop is a magnetic dipole with moment $IA$ along its axis. All $800$ turns face the same way and carry the current the same way round, so their moments add along the solenoid’s axis. The field outside the solenoid then has the same pattern as a bar magnet’s: the lines leave one end, which behaves as a north pole, and loop round into the other, a south pole. Suspended freely, the solenoid settles along the north–south line, and it attracts or repels a bar magnet exactly as another magnet would. Which end is north is fixed by the right-hand rule: curl the fingers of the right hand the way the current goes round, and the thumb points along $\vec m$, towards the north end.

Magnetic moment. The total moment is the sum over the turns:

$$m = NIA = (800)(3.0)(2.5\times10^{-4}) = 0.60\ \text{J T}^{-1}$$

Question 5.4

If the solenoid in Exercise 5.3 is free to turn about the vertical direction and a uniform horizontal magnetic field of $0.25\ \text{T}$ is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of $30^\circ$ with the direction of applied field?

Note on the question. The book prints “the solenoid in Exercise 5.5”, but Exercise 5.5 is about a bar magnet, and the only solenoid before this question is the one in Exercise 5.3. The book’s own answer, $7.5\times10^{-2}$, is what Exercise 5.3’s solenoid gives, so 5.3 is the exercise meant.

Solution. From Question 5.3 the solenoid has $m = 0.60\ \text{J T}^{-1}$, directed along its axis, so the angle between $\vec m$ and $\vec B$ is the $30^\circ$ between the axis and the field. The solenoid turns about a vertical axis and the field is horizontal, so the torque acts in the plane in which it is free to turn:

$$\tau = mB\sin\theta = (0.60)(0.25)\sin30^\circ = 7.5\times10^{-2}\ \text{N m}$$

Question 5.5

A bar magnet of magnetic moment $1.5\ \text{J T}^{-1}$ lies aligned with the direction of a uniform magnetic field of $0.22\ \text{T}$. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?

Question 5.5 (a)

Solution. If the external torque turns the magnet slowly, the magnet gains no kinetic energy, and all the work done goes into its potential energy. The work is therefore the change in $U = -mB\cos\theta$, from $\theta = 0^\circ$ to the final angle:

$$W = U_{\text{final}} – U_{\text{initial}} = -mB\cos\theta – (-mB\cos0^\circ) = mB(1 – \cos\theta)$$

Here $mB = (1.5)(0.22) = 0.33\ \text{J}$.

(i) Normal to the field, $\theta = 90^\circ$: $W = 0.33(1 – 0) = 0.33\ \text{J}$.

(ii) Opposite to the field, $\theta = 180^\circ$: $W = 0.33(1 – (-1)) = 0.66\ \text{J}$.

Question 5.5 (b)

Solution. The torque is $\tau = mB\sin\theta$, which depends on where the magnet ends up, not on the work done to get it there.

(i) At $\theta = 90^\circ$, $\sin\theta = 1$, so $\tau = mB = 0.33\ \text{N m}$, the largest it can be. It acts to turn $\vec m$ back towards $\vec B$.

(ii) At $\theta = 180^\circ$, $\sin\theta = 0$, so the torque is zero. This is the position of unstable equilibrium from Question 5.2(b): the energy is highest, but there is no torque until the magnet is disturbed.

Question 5.6

A closely wound solenoid of $2000$ turns and area of cross-section $1.6\times10^{-4}\ \text{m}^2$, carrying a current of $4.0\ \text{A}$, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5\times10^{-2}\ \text{T}$ is set up at an angle of $30^\circ$ with the axis of the solenoid?

Question 5.6 (a)

Solution. As in Question 5.3, the turns add their moments along the axis:

$$m = NIA = (2000)(4.0)(1.6\times10^{-4}) = 1.28\ \text{A m}^2$$

Question 5.6 (b)

Solution. Force. In a uniform field the net force on a dipole is zero. For a bar magnet the pull on its north pole is equal and opposite to the pull on its south pole; for a current loop the forces on opposite sides cancel in the same way. So the solenoid is not pushed anywhere, only turned.

Torque. The field is horizontal and the solenoid turns in a horizontal plane, so the torque acts in the plane it is free to turn in, with $\theta = 30^\circ$ between $\vec m$ and $\vec B$:

$$\tau = mB\sin\theta = (1.28)(7.5\times10^{-2})\sin30^\circ = 4.8\times10^{-2}\ \text{N m}$$

Question 5.7

A short bar magnet has a magnetic moment of $0.48\ \text{J T}^{-1}$. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of $10\ \text{cm}$ from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

The magnet is “short”, meaning its length is small compared with $10\ \text{cm}$, so it can be treated as a point dipole and the dipole field formulae apply, with $r = 0.10\ \text{m}$ and $r^3 = 10^{-3}\ \text{m}^3$.

Question 5.7 (a)

Solution. On the axis the field is twice as strong as on the equatorial line at the same distance:

$$B = \frac{\mu_0}{4\pi}\frac{2m}{r^3} = \frac{(10^{-7})(2\times0.48)}{10^{-3}} = 9.6\times10^{-5}\ \text{T} = 0.96\ \text{G}$$

On the axis the field points the same way as $\vec m$, that is, from the magnet’s south pole towards its north pole. At a point beyond the north pole this is simply the field leaving the north pole.

Note on the book’s answer. The key prints “0.96 g”. The lower-case g is a misprint: the unit is the gauss, $\text{G}$, as in part (b), and $0.96\ \text{G} = 9.6\times10^{-5}\ \text{T}$.

Question 5.7 (b)

Solution. On the equatorial line the formula has no factor of 2:

$$B = \frac{\mu_0}{4\pi}\frac{m}{r^3} = \frac{(10^{-7})(0.48)}{10^{-3}} = 4.8\times10^{-5}\ \text{T} = 0.48\ \text{G}$$

Here the field points opposite to $\vec m$, from north to south, because the field lines leaving the north pole curve round the side of the magnet on their way back to the south pole.

Common mistakes

• Question 5.1: using $\cos30^\circ$. The torque is $mB\sin\theta$, with $\theta$ the angle between the magnet’s axis and the field. The cosine belongs to the energy, $-mB\cos\theta$; using it here gives $0.21\ \text{J T}^{-1}$ instead of $0.36$.
• Question 5.2: getting the sign of the energy wrong. Stable equilibrium is the lowest energy, $-4.8\times10^{-2}\ \text{J}$, not zero and not the positive value. Students who think of energy as always positive, or put the zero of energy at the aligned position, mix up which orientation is stable.
• Question 5.4: taking “Exercise 5.5” literally. Using the $1.5\ \text{J T}^{-1}$ bar magnet of Exercise 5.5 gives $0.19\ \text{N m}$. The question is about a solenoid, and the solenoid is the one in Exercise 5.3, with $m = 0.60\ \text{J T}^{-1}$.
• Question 5.5(a): calculating the final energy, or the torque, instead of the change in energy. The work is $U_{\text{final}} – U_{\text{initial}} = mB(1 – \cos\theta)$. Leaving out the starting energy $-mB$ gives $0$ for case (i) and $0.33\ \text{J}$ for case (ii), both wrong.
• Question 5.5(b): expecting the largest torque when the magnet points against the field. That position has the most energy, but the torque there is zero; the torque is largest at $90^\circ$.
• Question 5.6(b): giving the solenoid a net force. A uniform field only turns a dipole. The forces on its two ends are equal and opposite, so the force is zero and only the torque, $4.8\times10^{-2}\ \text{N m}$, remains.
• Question 5.7: using the same formula on the axis and the equator, or mixing up their directions. The axial field has the factor $2$ and points along $\vec m$; the equatorial field has no $2$ and points opposite to $\vec m$. And $10\ \text{cm}$ cubed is $10^{-3}\ \text{m}^3$, not $10^{-2}$.

Practise next

• Chapter 6, Electromagnetic Induction — what happens when the flux of a magnet or a solenoid through a circuit changes, which is where the dipoles of this chapter start producing currents.
• Chapter 4, Moving Charges and Magnetism — $m = NIA$ and $\tau = NIAB\sin\theta$ come from the current loop there (Questions 4.9 and 4.13), and are worth revising together with 5.3 to 5.6.